Oh yeah, that's a usefull hint,
thank you very much Paul,
Georg
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Georg,
> is there an efficient way in sage to find the smallest integer k for
> which the inequality
>
> b^(k+1) / (factorial(k) * factorial(k+1)) <= 1
>
> is true (b > 0)
Stirling's expansion gives (when b goes to infinity) k ~ sqrt(b)*exp(1).
Thus it suffices to evaluates f(k) = b^(k+
Excuse me, i'm not a native english speaker (and i thought i read this
mode of speaking somewhere before):
is there an efficient way in sage to find the smallest integer k for
which the inequality
b^(k+1) / (factorial(k) * factorial(k+1)) <= 1
is true (b > 0)
similarly for
b^k / factorial(k)
On Jan 21, 2008 6:39 AM, Georg <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> is there an efficient way in sage to find the smallest integer k to
> meet (b constant)
>
> b^(k+1) / (factorial(k) * factorial(k+1)) <= 1
This sentence doesn't make sense to me. What does it mean for an
integer to meet an ine
